固体物理习题解答：Chapter 7 problems

1.Squarelattice,freeelectronenergies(a)Two-dimensionalsimplesquarelatticeAtthecornerofthefirstBrillouinzone:PointCkC=(/a,/a)AtthemidpointofasidefaceofthefirstBrillouinzone:PointAkA=(/a,0)OThereciprocallatticeofatwo-dimensionalsimplesquarelatticeisatwo-dimensionalsimplesquarelatticewiththelatticeconstant2/a.CBAThekineticenergyofafreeelectronintwodimensionsisThenandThus C=2A(b)Three-dimensionalsimplecubiclatticeThereciprocallatticeofathree-dimensionalsimplecubiclatticeisathree-dimensionalsimplecubiclatticewiththelatticeconstant2/a.AtthecornerofthefirstBrillouinzone:PointCkC=(/a,/a,/a)AtthemidpointofasidefaceofthefirstBrillouinzone:PointAkA=(/a,0,0)ThekineticenergyofafreeelectroninthreedimensionsisThenThus C=3A(c)DiscussionfortheconductivityofdivalentmetalsThereareNallowedelectronwavevectorsinthefirstBrilouinzone,whichallows2Nindependentorbitals.Foradivalentmetals,eachatomcontributestwovalenceelectrons.Nprimitivecellshave2Ntwovalenceelectrons.ThereforethevolumeoftheFermisphereisequaltothevolumeofthefirstBrilouinzone.i.e.IftheedgeofthesecondbandBislowerthanC,theelectronswillgointothesecondbandinsteadofthecornerofthefirstBrilouinzone.Forthedivalentmetals,theenergyofthesecondbandedgeissmallerthantheenergyatthecornerofthefirstBrilouinzonesothatanoverlapinenergyofthefirstandsecondbands.Therearevalenceelectronsinthesecondband,whichcontributetotheelectricconductivity.2.FreeelectronenergiesinreducedzoneThefreeelectronenergyofawavevectoriswhere isawavevectorinsidethefirstBrillouinzoneand isareciprocallatticevector.Thereciprocallatticeofanfcccrystallatticeisabcclatticewithlatticeconstant4/a.Inthe[111]direction,InthefirstBrillouinzone,1/2

l1/2.Thereciprocallatticevectorwhereu,v,andwareintegers.withTherefore(1)Forthefirstenergyband,u=v=w=0,AtthefirstBrillouinzonebourdary(pointL),l=1/2.Thebottomoftheenergyband(kx,ky,kz,)=(0,0,0),

(0,0,0)=0.Inthe[111]direction(2)Forthe2ndand3rdenergybands,u=v=w=1,At(kx,ky,kz,)=(0,0,0),Inthe[111]direction(3)Forthe4th,5th,and6thenergybandsandu=v=0,w=1,andv=w=0,u=1,At(kx,ky,kz,)=(0,0,0),u=w=0,v=1,Inthe[111]direction(4)Forthe7th,8th,and9thenergybandsandu=v=0,w=1,andv=w=0,u=1,At(kx,ky,kz,)=(0,0,0),u=w=0,v=1,Inthe[111]directionv=0,u=w=1,andw=0,u=v=1,At(kx,ky,kz,)=(0,0,0),andu=0,v=w=1,(5)Forthe10th,11th,and12thenergybandsInthe[111]directionv=0,u=w=1,andw=0,u=v=1,At(kx,ky,kz,)=(0,0,0),andu=0,v=w=1,(6)Forthe13th,14th,and15thenergybandsInthe[111]directionwhere and03.Kronig-PenneymodelForthedelta-functionpotential,theKronig-Penneymodelgives(a)Atk=0,SetthenThenForthelowestenergyband,n=0.InthelimitP<<1,ThenweobtainTheenergyofthelowestenergybandis(b)Atk=/a,Withthesameprocess,wehaveThenForthelowestenergyband,n=0.InthelimitP<<1,FromEq(2),wehaveThebandgapatk=/aisThesolution isnotcorrectbecauseitgivestheenergyclosetothebottomoftheenergyband.4.Potentialenergyinthediamondstructure(a)Therearetwoatomsinthebasisofthediamondstructure: (0,0,0)and(a/4,a/4,a/4).ThenthecrystalpotentialmaybewrittenaswithwhereIfisabasisvectorinthereciprocallatticereferredtotheconventionalcubiccell,thenThediamondstructureisafcccrystallattice.Itsreciprocallatticeisabcclatticewiththelatticeconstant4/a.Therefore(b)AttheBrillouinzoneboundary,k=G/2.Inthefirst-orderapproximation,onlyCG/2areconsidered.Thecenterequationbecomes(page190)Aswecalculatedabove,U=UG=0.Then=,andtheenergygapvanishesatthezoneboundary.Inahigherorderapproximationwewouldgobacktothecenterequationswhereanynon-vanishingUGenters.5.ComplexwavevectorsintheenergygapAttheBrillouinzoneboundary,wehavethewavevectoratthecenteroftheenergygapisThenThesecularequation(page190Eq.(46))becomes:wherek<<G.AtthecenteroftheenergygapThus6.SquarelatticeThecrystalpotentialofasquarelattice